31.

Mr. Richard has a recurring deposit account in a bank for 3 years at 7.5%  p. a. simple interest. If he gets Rs. 8325  as interest at the time of maturity, find 

(i) The monthly deposit

(ii) The maturity value

( i )  Let the deposit per month  = Rs. P

Number of months = ( n ) = 36

Rate of interest ( r ) = 7.5 % p.a.

$$\begin{gathered} \therefore \mathrm{S}.\mathrm{I}. = \mathrm{P} \mathrm{x} \frac{ \mathrm{n} \mathrm{x} ( \mathrm{n} + 1 )}{2 \mathrm{x} 12} \mathrm{x} \frac{\mathrm{r}}{100} \\ \Rightarrow 8325 = \mathrm{P} \mathrm{x} \frac{36 \mathrm{x} ( 36 + 1 )}{2 \mathrm{x} 12} \mathrm{x} \frac{7.5}{100} \\ \Rightarrow 8325 = \mathrm{P} \mathrm{x} \frac{36 \mathrm{x} 37}{2 \mathrm{x} 12} \mathrm{x} \frac{7.5}{100} \\ \Rightarrow 8325 = \mathrm{P} \mathrm{x} \frac{3 \mathrm{x} 37}{2 } \mathrm{x} \frac{7.5}{100} \\ \Rightarrow \mathrm{P} = \frac{8325 \mathrm{x} 2 \mathrm{x} 100}{3 \mathrm{x} 37 \mathrm{x} 7.5} \\ \Rightarrow \mathrm{P} = \mathrm{Rs}. 2000 \end{gathered}$$

( ii ) Maturity value = P x n + S.I. 

                            = Rs. ( 2000 x 36 + 8325 )

                            = Rs. 80,325.